\section{The two-normal transfer with a fractional trace}
\label{sec:fractional-two-normal}

When the trace has a nonzero Euler kernel, attaching every residual
point to a trace point preserves that kernel.  We therefore leave one
residual projection free.  The extra point detects the trace kernel,
while the lower-degree interpolation estimates still control the
normal part of every relation.

\begin{proposition}\label{prop:fractional-two-normal}
Let \(D\geq4\), \(m\geq D(D-1)\), and \(n=m+2\).  Assume ideal
generation in degrees \(D-2\) and \(D-1\) in every dimension, and in
degree \(D\) for general points of \(\mathbf P^{m-1}\).  If
\(E(m,D)\notin\mathbf Z\), then ideal generation in degree \(D\)
holds for general points of \(\mathbf P^{n-1}\).
\end{proposition}

\begin{proof}
We prove maximal rank at each of the two critical counts.  Retain the
notation
\[
 \begin{gathered}
 T=k[y_1,\ldots,y_m],\qquad S=T[s,t],\qquad e=D-2,\\
 F=N_D(m),\quad M=N_{D-1}(m),\quad L=N_{D-1}(m+1),\\
 C=N_e(n),\qquad R=N_e(m+1),\qquad M_e=N_e(m).
 \end{gathered}
\]
In particular,
\begin{equation}\label{eq:fractional-binomial-identities}
 N_D(n)-F=2L+C,\qquad DF=mL,\qquad (m+1)C=(D-1)L.
\end{equation}
Write
\[
 r_2=\lceil E(m,D)\rceil,\quad
 \delta=r_2-E(m,D),\quad p_2=F-r_2.
\]
Thus \(0<\delta<1\).  The Euler evaluation on \(p_2\) general trace
points is surjective and its kernel has dimension
\(\kappa=(m-1)\delta\).
For either \(r=\lfloor E(n,D)\rfloor\) or
\(r=\lceil E(n,D)\rceil\), put
\[
 \Delta=r-E(n,D),\qquad
 p=N_D(n)-r,\qquad p_1=p-p_2,\qquad \sigma=(m+1)\Delta.
\]
The integer \(\sigma\) is the source dimension minus the target
dimension of the ambient Euler evaluation.

\smallskip
\noindent\emph{The numbers of residual points.}
Choose
\begin{equation}\label{eq:fractional-counts}
 \begin{gathered}
 b=\left\lceil\frac{DL}{D+1}+\frac{\delta}{m}\right\rceil+1,
 \qquad \theta=b-\frac{DL}{D+1},\qquad a=m\theta-\delta,\\
 q=p_1-2b,\qquad
 g=C-q=\frac{C}{D+1}-\delta+\Delta+2\theta .
 \end{gathered}
\end{equation}
If \(\eta\in[0,1)\) denotes the rounding error in the ceiling, then
\[
 \theta=1+\frac{\delta}{m}+\eta,\qquad
 a=m(1+\eta),\qquad m\leq a<2m.
\]
Consequently
\begin{equation}\label{eq:fractional-positive-kernel}
 d:=L+(m+1)g-2b=2a+\sigma-\kappa>0,
\end{equation}
because \(\sigma>-(m+1)\) and \(\kappa<m-1\).  The additional
one in the definition of \(b\) ensures this positivity.

Here are the bounds needed below:
\begin{equation}\label{eq:fractional-needed-bounds}
 \begin{gathered}
 R\leq b<M,\qquad L-b<\frac{L}{D},\qquad
 0<g+1<\frac{M_e}{D-1},\\
 b>(m+1)(g+1),\qquad 1<q<C,\\
 p_1-1\leq p_2,\qquad p_2\geq M.
 \end{gathered}
\end{equation}
We verify them explicitly.  First, the endpoint
\(m_0=D(D-1)\) has integral \(E(m_0,D)\).  Indeed,
\[
 D\binom{D^2-1}{D}=(D^2-1)\binom{D^2-2}{D-1},
\]
and \(D\) is coprime to \(D+1\).  We may therefore assume
\(m\geq m_0+1\geq13\); in particular \(D+1\leq m/2\).
Since \(\theta<2+\delta/m\) and \(|\Delta|<1\),
\[
 0<g<\frac{C}{D+1}+5,\qquad
 g+1<\frac{C}{D+1}+6.
\]
The margin for the preceding degree is
\begin{equation}\label{eq:fractional-margin}
 \frac{M_e}{D-1}-\frac{C}{D+1}
 =\frac{M_e}{m(m+1)}
 \left\{\frac{m(m+1)}{D-1}
       -\frac{(m+D-2)(m+D-1)}{D+1}\right\}.
\end{equation}
The expression in braces is increasing for \(m\geq m_0\), and
equals \(3D-2\) at \(m=m_0\); the factor preceding it is
nondecreasing and at least \(1/2\).  For \(D\geq5\) the margin is
therefore at least \((3D-2)/2>6\).  For \(D=4\) it is exactly
\[
 \frac{m^2-5m-9}{15}\geq\frac{19}{3}>6
 \qquad(m\geq13).
\]
This proves the bound for \(g+1\).  Its associated trace point
count is in the degree-\(e\) initial cell: for \(m\geq2\),
\[
 \frac{M_e}{e+1}\leq M_e-N_{e-1}(m).
\]

Next,
\[
 M-\frac{DL}{D+1}
 =\frac{M(m-m_0)}{m(D+1)}
 \geq\frac{(m+1)(m+2)}{6(D+1)}
 \geq\frac{(m+1)(m+2)}{3m}>4.
\]
Here \(M\geq N_3(m)\), while \(\theta<3\).  Hence \(b<M\).
The ratio \(L/R=(m+D-1)/(D-1)\) also gives \(b\geq DR\geq R\).
Thus \(b\) belongs to the degree-\((D-1)\) initial cell and
\(L-b\leq L/(D+1)<L/D\), as required for the lower-degree
injectivity hypothesis.

For the pure normal spaces, substitution in
\eqref{eq:fractional-counts} gives
\[
 b-(m+1)(g+1)>\frac{L}{D+1}-6m-4>0.
\]
For the second inequality use \(L\geq N_3(m+1)\) and
\(D+1\leq m/2\).  After multiplication by \(3m\), the resulting
lower bound has numerator
\(m^3-12m^2-m+6\), which is positive for \(m\geq13\).
Moreover,
\[
 q>\frac{DC}{D+1}-5\geq\frac{4C}{5}-5>1,
\]
since \(C\geq N_2(n)\geq120\); positivity of \(g\) gives \(q<C\).

Finally,
\[
 \begin{split}
 p_2-p_1+1
 &=\frac{(m-2D)L-DC}{D+1}-2\delta+\Delta+1\\
 &>\frac{L}{D+1}
       \left(m-2D-\frac{D(D-1)}{m+1}\right)-2>0.
 \end{split}
\]
Indeed, \(m-2D\geq D(D-3)\geq4\),
\(D(D-1)/(m+1)<1\), and \(L>D+1\).
Also
\[
 p_2=\frac{mL}{D+1}-\delta\geq(D-1)M-\delta\geq M.
\]
This completes \eqref{eq:fractional-needed-bounds}.
Both \(p_2\) and \(p_2+1\) lie in the degree-\(D\) trace cell;
the corresponding ideal dimensions are respectively the upper
and lower critical integers.  The same binomial ratio places
the two ambient critical counts in their degree-\(D\) cell.

\smallskip
\noindent\emph{The marked configuration and its slice evaluations.}
Set
\[
 \Lambda=\{s=t=0\},\qquad H_A=\{t=0\},\qquad H_B=\{s=0\}.
\]
Choose \(b\) general points \(A\) on \(H_A\setminus\Lambda\),
\(b\) general points \(B\) on \(H_B\setminus\Lambda\), and \(q\)
general points \(X\) with both \(s\) and \(t\) nonzero.  Every
point has nonzero \(y\)-coordinates.  Designate one point
\(x_*\in X\), and let \(y_*\in\Lambda\) be its projection.
The trace set \(Z_2\) contains the distinct projections of
\(A\), \(B\), and \(X\setminus\{x_*\}\), together with enough
additional general points to have cardinality \(p_2\).
The inequality \(p_1-1\leq p_2\) makes this possible.  The
projection \(y_*\) is general independently of the ordered
trace set.  Write
\[
 Z=Z_2\cup A\cup B\cup X.
\]

These choices form an irreducible parameter space.  Its map to
the ordered pairs \((Z_2,y_*)\) is dominant: the trace projections
may be prescribed first, and all normal coordinates then vary
freely.  We impose below finitely many nonempty rank-open
conditions on this space.  In particular, all uses of general
points refer to a common choice of \(Z\).

Let
\[
 \mathcal T=\ker\!\left(
 H^0(\Lambda,\Omega_\Lambda(D+1))
 \longrightarrow
 H^0(Z_2,\Omega_\Lambda(D+1)|_{Z_2})\right).
\]
The trace hypothesis gives \(\dim\mathcal T=\kappa\) and
surjectivity of this evaluation.  A section
\(\tau=(\tau_i(y))\in\mathcal T\) extends to
\((\tau_i(y),0,0)\) in the ambient Euler bundle.  This extension
vanishes on both \(A\) and \(B\), because their projections
belong to \(Z_2\).

On \(H_A\), put \(W_A=I_A(D-1)\) and
\[
 U_A=\sum_{i=1}^m y_i(W_A|_\Lambda)\subset T_D;
\]
define \(W_B,U_B\) similarly.  The preceding degree gives
injective multiplication on \(W_A\) and \(W_B\).
Lemma~\ref{lem:attached-code} applies, since \(b<M\).  It gives
\(\dim U_A=\dim U_B=m(L-b)\), and evaluation on every subset
of \(m(L-b)\) trace labels is an isomorphism.

The kernel of the slice Euler evaluation on \(Z_2\cup A\) is
exactly the extended space \(\mathcal T\).  Indeed, restrict a
slice kernel section to its tangent trace component and subtract
the corresponding extension of \(\tau\).  Its tangent trace
components now vanish as polynomials.  The remaining calculation
is the one used in the integral case: divide those components
by \(s\), use injectivity of \(U_A\) on \(Z_2\), and then use
injectivity of linear multiplication on \(W_A\).  The remainder
is zero.  The same argument holds on \(H_B\).

The scalar descriptions of the slice cokernels are consequently
\begin{equation}\label{eq:fractional-slice-cokernels}
 Q_A=k^{Z_2}/\operatorname{ev}_{Z_2}(U_A),\qquad
 Q_B=k^{Z_2}/\operatorname{ev}_{Z_2}(U_B),
 \qquad \dim Q_A=\dim Q_B=a.
\end{equation}
To see the cokernel description, first remove the tangent
trace values by a trace Euler section, and then remove the
values at \(A\) by a section whose tangent components are
divisible by \(s\).  The remaining scalar trace values are
taken modulo \(U_A\).  The possible ambiguity from \(\mathcal T\)
contributes zero, because its displayed extension vanishes
at \(A\).  This proves \eqref{eq:fractional-slice-cokernels};
the argument on \(B\) is identical.  The interpolation statement
for \(U_A,U_B\) says that every \(a\) coordinate classes form a
basis of the corresponding \(Q_A,Q_B\).

\smallskip
\noindent\emph{Surjectivity of the normal part at \(X\).}
Consider the space \(V\) of Euler sections
\begin{equation}\label{eq:fractional-axis-form}
 (st a_1,\ldots,st a_m,t a_s,s a_t),
 \qquad \sum_i y_i a_i+a_s+a_t=0,
\end{equation}
that vanish at \(A\cup B\).  Here \(a_i\in S_e\) and
\(a_s,a_t\in S_{D-1}\).  Before imposing \(A,B\), the free
parameters are \((a_1,\ldots,a_m,a_s)\), so their dimension is
\((m+1)C+L\).  The remaining conditions are
\[
 (a_s+\textstyle\sum_i y_i a_i)|_A=0,\qquad a_s|_B=0.
\]
The map to the two complete slice polynomial spaces is onto.
For any prescribed pair, choose \(a_s\) with the desired
restriction on \(H_B\); the difference of the desired
restriction on \(H_A\) and \(a_s|_{H_A}\) lies in the ideal
\((y_1,\ldots,y_m)\), apart from its possible pure \(s\)-term,
which can first be incorporated in \(a_s\).
The remaining difference is \(\sum y_i a_i\).
Thus the \(2b\) scalar conditions are independent, and
\[
 \dim V=(m+1)C+L-2b.
\]

Put \(G=I_X(e)\).  The height degeneration of
Lemma~\ref{lem:height-degeneration}, applied also to
\(\widetilde G=I_{X\setminus\{x_*\}}(e)\), gives
\begin{equation}\label{eq:fractional-height-ranks}
 \begin{gathered}
 \dim G=g,\qquad\dim\widetilde G=g+1,\\
 \widetilde G\longrightarrow T_e\ \text{injective},\qquad
 (\widetilde G|_\Lambda)\otimes T_1\longrightarrow T_{e+1}
 \ \text{injective}.
 \end{gathered}
\end{equation}
The numerical hypothesis is \(g+1<M_e/(e+1)\), already
proved above, and the required ordinary point ideal is in
its initial cell.
The same restriction and multiplication injections hold
for the subspace \(G\).

The kernel of evaluation \(V\to\bigoplus_{x\in X}
\Omega_{\mathbf P^{n-1}}(D+1)|_x\) is described by
\[
 (a_i)\in G^m,\qquad a_s\in I_X(D-1),
\]
subject to the two scalar conditions at \(A,B\).
As in the integral case, the associated polynomial-pair space
\[
 (a_i,a_s)\longmapsto
 \left((a_s+\textstyle\sum_i y_i a_i)|_{H_A},\,
                         a_s|_{H_B}\right)
\]
has dimension \(L+(m+1)g\), and each of its pure-side
subspaces has dimension \((m+1)g\).
Its parametrization is injective by
\eqref{eq:fractional-height-ranks}: vanishing of both outputs
first forces all \(a_i=0\) on restriction to \(\Lambda\),
then makes \(a_s\) divisible by \(st\), and restriction
injectivity for \(G\) kills the remaining factor.
Lemma~\ref{lem:two-group-evaluation} and
\(b>(m+1)g\) now give kernel dimension
\[
 L+(m+1)g-2b=d>0.
\]
It follows that this evaluation on \(X\) is surjective, because
\[
 \dim V-d=(m+1)(C-g)=(m+1)q
\]
is its target dimension.

Every ambient kernel section has a unique decomposition into
an extended \(\tau\in\mathcal T\) and a section of the form
\eqref{eq:fractional-axis-form}.  Surjectivity just proved
allows us to compensate the values of every \(\tau\) at \(X\).
Thus, if \(K\) is the full Euler kernel on \(Z\),
\[
 \dim K=\kappa+d=2a+\sigma.
\]
Its cokernel has dimension \(2a\).
Restriction to the two slices gives a surjection from this
cokernel to \(Q_A\oplus Q_B\): tangent trace representatives
can be aligned using surjectivity of the trace evaluation,
and the two normal trace values are independent.  The
dimensions agree, so
\begin{equation}\label{eq:fractional-ambient-cokernel}
 \operatorname{coker}(\operatorname{ev}_Z)
 \simeq Q_A\oplus Q_B.
\end{equation}

\smallskip
\noindent\emph{The free projection detects the trace kernel.}
We show that each of the two maps
\begin{equation}\label{eq:fractional-scalar-detection}
 K\longrightarrow T_{D-1},\qquad
 F\longmapsto a_s|_\Lambda,\quad
 \text{respectively }F\longmapsto a_t|_\Lambda,
\end{equation}
is injective.
At every point of \(X\setminus\{x_*\}\), the extension of
\(\tau\) vanishes by attachment.  Hence all \(a_i\) lie in
\(\widetilde G\).  Its normal components vanish everywhere,
so \(a_s\) and \(a_t\) vanish at all of \(X\).

Suppose \(a_s|_\Lambda=0\).  On \(H_B\), write
\(a_s=t h_B\), with \(\deg h_B=e\).  Since \(t\ne0\) on
\(B\), the polynomial \(h_B\) vanishes there.  The bound
\(b\geq R\) and general scalar interpolation force \(h_B=0\).
Thus \(a_s\) vanishes identically on \(H_B\), so
\(a_s=s h\) in \(S\).  As \(s\ne0\) on \(X\), we have
\(h\in G\subset\widetilde G\).

The condition at \(A\) is now
\[
 \left(\sum_i y_i a_i+s h\right)|_A=0.
\]
The polynomial space on \(H_A\) parametrized by
\((a_1,\ldots,a_m,h)\in\widetilde G^{m+1}\) has dimension
\((m+1)(g+1)\).  Indeed, if its polynomial is zero, restriction
to \(\Lambda\) and \eqref{eq:fractional-height-ranks} give
all \(a_i=0\).  Then \(h|_{H_A}=0\), hence \(h|_\Lambda=0\);
restriction injectivity for \(\widetilde G\) gives \(h=0\).
General scalar
evaluation at the \(b>(m+1)(g+1)\) points of \(A\) is injective
on this space.  It follows that all \(a_i,h\), and hence
the whole normal section, vanish.

Only \(\tau\) remains.  Its vanishing at \(x_*\) says that
\(\tau(y_*)=0\).  The trace set \(Z_2\cup\{y_*\}\) has ideal
dimension
\[
 r_2-1=\lfloor E(m,D)\rfloor.
\]
The injective trace edge therefore gives \(\tau=0\).
This proves the first injection in
\eqref{eq:fractional-scalar-detection}; interchanging \(A,s\)
with \(B,t\) proves the second.

The scalar rank conditions in this argument can all be
imposed on the same marked family.  For example, fix
\(X\setminus\{x_*\}\) satisfying
\eqref{eq:fractional-height-ranks}, and then choose \(A\)
and \(B\) to give the required scalar injections.  These
are nonempty opens on the full parameter space.  The
trace conditions for \(Z_2\) and \(Z_2\cup\{y_*\}\) are
nonempty opens under its dominant map to \((Z_2,y_*)\).
Their intersection with the other rank opens is nonempty.
In particular, no independence of an attached point from
its own projection is required.

\smallskip
\noindent\emph{Normal motion and maximal rank.}
Move the original trace points, keeping their \(y\)-coordinates
fixed, by
\[
 s(z)=\varepsilon u_z,\qquad t(z)=\varepsilon v_z
 \qquad(z\in Z_2),
\]
and keep \(A,B,X\) fixed.  The extension of \(\tau\) has
zero derivative in these directions.  The first derivative
of \eqref{eq:fractional-axis-form}, followed by
\eqref{eq:fractional-ambient-cokernel}, is therefore
\begin{equation}\label{eq:fractional-first-response}
 J=\sum_{z\in Z_2}v_z e_z^A\otimes
        [F\mapsto a_s|_\Lambda(z)]
   +\sum_{z\in Z_2}u_z e_z^B\otimes
        [F\mapsto a_t|_\Lambda(z)].
\end{equation}
Each family of source functionals spans \(K^*\), by
\eqref{eq:fractional-scalar-detection} and \(p_2\geq M\).
Every \(a\) coordinate classes in either target form a
basis. Lemma~\ref{lem:two-normal-response} therefore gives a
choice of normal velocities for which \(J\) has maximal rank,
and the deformed Euler evaluation has the required rank at
the chosen critical count.

For completeness, \(Z\) has independent degree-\(D\) scalar
evaluation.  Interpolate its trace values in \(T_D\), then
correct the values at \(A\) and \(B\) separately with \(s\)
and \(t\) times slice forms of degree \(D-1\).  Finally
correct the values at \(X\) with \(stS_e\); evaluation of
\(S_e\) at \(X\) is onto because \(q<C\) and \(X\) is general.
A linear form avoiding \(Z\) gives degree-\((D+1)\)
independence as well.  These properties and distinctness
persist under the deformation.  The Euler criterion
identifies the resulting maximal rank with that of the
point-ideal multiplication map.  Doing this at the two
critical counts, followed by point addition and deletion,
proves the proposition.
\end{proof}